I am stuck on this question and not really sure what to do for it. I think that i can use this equation Q/dt=e*sigma*A*T^4 for it, but i'm not getting the right answer for it. Is this not the right equation?
Question:
Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does using thin, 1.8 -by-3.6 panels that have a working temperature of about 6. How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible.
Hint: Don't forget that the panels have two sides!
I assume you are using radiating temperature as 6K.
Now, on absorbed radiation. Do you mean from the Sun (in the shade)? But in the shade, the background radiation is about 2.3K. I guess I am not certain what you are doing.
To calculate the power radiated from each panel, you can use the Stefan-Boltzmann law, which is the correct equation for this situation. The equation you mentioned, Q/dt = e * sigma * A * T^4, is indeed the correct form of the Stefan-Boltzmann law, where Q/dt represents the power (rate of energy transfer), e is the emissivity of the surface, sigma is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature in Kelvin.
However, in this specific question, there are a few additional details to consider. The panels have two sides, meaning that both sides will contribute to the total power radiated. Additionally, the panels are in the shade, which means that they won't be affected by any absorbed radiation.
Let's break down the steps to solve this question:
1. Convert the temperature provided to Kelvin:
Working temperature = 6 °C + 273.15 = 279.15 K
2. Calculate the power radiated from each side of the panel separately:
Power = 2 * e * sigma * A * T^4
3. The emissivity for most materials is approximately 0.9, so you can assume an emissivity of 0.9 for the panels.
4. The surface area of each panel is given as 1.8 cm * 3.6 cm. However, it's important to convert it to square meters to match the units of the other variables in the equation. Use the conversion factor:
1 cm^2 = (1 cm) * (1 cm) = (0.01 m) * (0.01 m) = 0.0001 m^2
Now, multiply the given surface area by the conversion factor to get the area in square meters.
5. Plug in the values into the equation and calculate the power:
Power = 2 * 0.9 * sigma * A * T^4
6. Finally, substitute the values of the variables into the equation, making sure to use Kelvin for temperature:
Power = 2 * 0.9 * (5.67 × 10^-8 W/(m^2*K^4)) * (Area in square meters) * (279.15 K)^4
Solving this equation will give you the power radiated from each panel.